\fuxiti
\begin{enhancedline}
\begin{xiaotis}

\xiaoti{$x$ 是什么值的时候，下列各式在实数范围内才有意义？}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={12em, colsep=0pt}}
        \xxt{$\sqrt{x - 3}$；} & \xxt{$\sqrt{3 - x}$；} & \xxt{$\sqrt{1 + x^2}$；} \\
        \xxt{$\sqrt{\dfrac{1}{x^2}}$；} & \xxt{$\sqrt{x} + \sqrt{-x}$；} & \xxt{$\dfrac{1}{1 - \sqrt{x}}$。}
    \end{tblr}
\end{xiaoxiaotis}


\xiaoti{在实数范围内把下列各多项式分解因式：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}} %, rows={rowsep=0.5em}}
        \xxt{$x^2 - 7$；} & \xxt{$4a^4 - 1$；} \\
        \xxt{$a^4 - 6a^2 + 9$；} & \xxt{$m^4 - 10m^2n^2 + 25n^4$。}
    \end{tblr}
\end{xiaoxiaotis}

\xiaoti{下面的推导错在哪里？}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={colsep=0pt}}
        \xxt{$\begin{tblr}[t]{}
            \because    & (-3)^2 = 3^2 \douhao \\
            \therefore  & \sqrt{(-3)^2} = \sqrt{3^2} \juhao \\
            \because    & \sqrt{(-3)^2} = -3 \douhao \quad \sqrt{3^2} = 3 \douhao \\
            \therefore  & -3 = 3 \juhao
        \end{tblr}$} & \xxt{$\begin{tblr}[t]{}
            \because    & -2\sqrt{3} = \sqrt{(-2)^2 \times 3} = \sqrt{12} \douhao \\
            \text{而}   & \sqrt{12} = 2\sqrt{3} \douhao \\
            \therefore  & -2\sqrt{3} = 2\sqrt{3} \juhao \\
            \therefore  & -2 = 2 \juhao
        \end{tblr}$}
    \end{tblr}
\end{xiaoxiaotis}


\xiaoti{什么叫做最简二次根式？把下列各式化成最简二次根式：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}, rows={rowsep=0.5em}}
        \xxt{$\sqrt{500}$；} & \xxt{$\sqrt{4\dfrac{2}{3}}$；} \\
        \xxt{$\sqrt{12x}$；} & \xxt{$\sqrt{3a^2b^2} \quad (b < 0)$；} \\
        \xxt{$\sqrt{\dfrac{2}{3ab^2}}$；} & \xxt{$x^2\sqrt{\dfrac{y}{8x}}$；} \\
        \xxt{$\sqrt{\dfrac{x^2 - y^2}{a}} \quad (x > y)$；} & \xxt{$(x - y)\sqrt{\dfrac{b^3}{x^2 - y^2}} \quad (x > y)$；} \\
        \xxt{$\sqrt{(a^2 - b^2) (a^4 - b^4)} \quad (a > b)$；} & \xxt{$\sqrt{a^{2n+1}b^3}$。}
    \end{tblr}
\end{xiaoxiaotis}


\xiaoti{什么叫做同类二次根式？下列二次根式里，哪几个是同类二次根式？\\
    $\sqrt{44}$\nsep  $\sqrt{\dfrac{1}{x}}$\nsep  $-\sqrt{1\dfrac{5}{11}}$\nsep
    $\sqrt{x^3y^2}$\nsep  $\sqrt{175}$\nsep  $2\sqrt{a^2x}$\nsep
    $\dfrac{1}{2}\sqrt{63}$\nsep \\
    $-\sqrt{99}$\nsep  $5\sqrt{3\dfrac{4}{7}}$\nsep
    $\sqrt{\dfrac{m}{1 - 2x + x^2}} \quad (x > 1)$\nsep
    $\sqrt{225m^3}$。
}

\xiaoti{计算：}
\begin{xiaoxiaotis}

    \xxt{$\left(\sqrt{24} - \sqrt{\dfrac{1}{2}} + 2\sqrt{\dfrac{2}{3}}\right) - \left(\sqrt{\dfrac{1}{8}} + \sqrt{6}\right)$；}

    \xxt{$7\sqrt{a} + 5\sqrt{a^2x} - 4\sqrt{\dfrac{b^2}{a}} - 6\sqrt{\dfrac{b^2x}{9}}$；}

    \xxt{$2\sqrt{12} \cdot \dfrac{1}{4}\sqrt{3} \div 5\sqrt{2}$；}

    \xxt{$9\sqrt{45} \div 3\sqrt{\dfrac{1}{5}} \times \dfrac{3}{2}\sqrt{2\dfrac{2}{3}}$；}

    \xxt{$\left(6\sqrt{\dfrac{3}{2}} - 5\sqrt{\dfrac{1}{2}}\right) \left(\dfrac{1}{4}\sqrt{8} - \sqrt{\dfrac{2}{3}}\right)$；}

    \xxt{$\sqrt{2x} - 3\sqrt{8x^3} \div 8\sqrt{\dfrac{x}{4}}$；}

    \xxt{$(10\sqrt{48} - 6\sqrt{27} + 4\sqrt{12}) \div \sqrt{6}$；}

    \xxt{$(2\sqrt{3} + 3\sqrt{6}) (2\sqrt{3} - 3\sqrt{6})$；}

    \xxt{$(\sqrt{x} + \sqrt{x - 1}) (\sqrt{x} - \sqrt{x - 1}) \quad (x > 1)$；}

    \xxt{$(8\sqrt{5} + 6\sqrt{3})^2$；}

    \xxt{$\left(\dfrac{3}{2}\sqrt{1\dfrac{2}{3}} - \sqrt{1\dfrac{1}{4}}\right)^2$；}

    \xxt{$(\sqrt{2} + 2\sqrt{3} - 3\sqrt{6}) (\sqrt{2} - 2\sqrt{3} + 3\sqrt{6})$；}

    \xxt{$\dfrac{\sqrt{5}}{\sqrt{3} + 1} + \dfrac{\sqrt{3}}{\sqrt{5} - \sqrt{3}} - \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}$；}

    \xxt{$\dfrac{4\sqrt{5} + 3\sqrt{6}}{3\sqrt{5} - 2\sqrt{6}} + \dfrac{\sqrt{5}}{\sqrt{5} + \sqrt{2}}$；}

    \xxt{$(5\sqrt{3} + 2\sqrt{5}) \div (2\sqrt{3} - \sqrt{5})$；}

    \xxt{$(\sqrt{2} + \sqrt{3} + \sqrt{5}) (3\sqrt{2} + 2\sqrt{3} - \sqrt{30})$；}

    \xxt{$\dfrac{n + 2 + \sqrt{n^2 - 4}}{n + 2 - \sqrt{n^2 - 4}} + \dfrac{n + 2 - \sqrt{n^2 - 4}}{n + 2 + \sqrt{n^2 - 4}} \quad (n > 2)$。}

\end{xiaoxiaotis}

\xiaoti{当 $x = 2 - \sqrt{3}$ 时，求代数式 \\
    \hspace*{2em} $(7 + 4\sqrt{3})x^2 + (2 + \sqrt{3})x + \sqrt{3}$ 的值。
}

\xiaoti{已知 $x = \dfrac{1}{2}(\sqrt{7} + \sqrt{5})$，
    $y = \dfrac{1}{2}(\sqrt{7} - \sqrt{5})$，求下列各式的值：
}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$x^2 - xy + y^2 $；} & \xxt{$\dfrac{x}{y} + \dfrac{y}{x}$。}
    \end{tblr}
\end{xiaoxiaotis}


\xiaoti{已知 $x_1 = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}$，
    $x_2 = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}$，其中 $a$，$b$，$c$ 都是实数，
    并且 $b^2 - 4ac \geqslant 0$，试计算下列各式的值：
}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}} %, rows={rowsep=0.5em}}
        \xxt{$x_1 + x_2$；}         & \xxt{$x_1 \cdot x_2$；} \\
        \xxt{$ax_1^2 + bx_1 + c$；} & \xxt{$ax_2^2 + bx_2 + c$。}
    \end{tblr}
\end{xiaoxiaotis}


\xiaoti{解下列方程：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}} %, rows={rowsep=0.5em}}
        \xxt{$\sqrt{6}(x + 1) = \sqrt{7}(x - 1)$；}
            & \xxt{$\dfrac{\sqrt{3}x}{\sqrt{2}} + 1 = \dfrac{2\sqrt{2}x}{\sqrt{3}}$。}
    \end{tblr}
\end{xiaoxiaotis}


\xiaoti{解下列方程组：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={19em, colsep=0pt}}
        \xxt{$\begin{cases}
                \sqrt{3}x - \sqrt{2}y = 1 \douhao \\
                \sqrt{2}x - \sqrt{3}y = 0 \fenhao
            \end{cases}$}
          & \xxt{$\begin{cases}
                \sqrt{2}x + \sqrt{3}y = \sqrt{7} \douhao \\
                \sqrt{6}x - \sqrt{7}y = \sqrt{5} \juhao
            \end{cases}$}
    \end{tblr}
\end{xiaoxiaotis}

\end{xiaotis}
\end{enhancedline}
